# Minimum no of swaps required to sort the array

Minimum no of swaps required to sort the array

hi @kumarakash121005
refer this -->

u will get very good explanation with codeâ€¦

Can I solve it without graph

Not getting logic of the problem

hi @kumarakash121005
yes u can do it without graphs alsoâ€¦ scroll down the link, u will get other approach alsoâ€¦

Is this problem same as selection sort

noâ€¦ not selection sort as suchâ€¦

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@kumarakash121005 this problem can be solved using 1) Graphs 2) hashing.
Since you are more concerned about the non-graph approach I will try to explain to you the Hashmap approach.

We will store elements in the array as a pair of values and their index positions as keys.

1. Sort the given array based on their values. Please Note we could include duplicate values as well. So if the current elementâ€™s value in the sorted array is equal to the element or the index is equal to the hashed index position in the original array. No swap is needed, and we can move to the next iteration.

2. But if the above condition doesnâ€™t hold, we will swap the element, say at ith index element of the array with the hashed index element in the array.

3. Keep on doing this until we donâ€™t satisfy the above criterion (1).

4. Now increment the answer.

Reffer to below code for more understanding of the approach

#include <bits/stdc++.h>
using namespace std;

int findMinSwap(vector &arr, int n)
{
// temporary vector to store values, along with its index in the original vector
vector<pair<int, int>> temp(n);
for (int i = 0; i < n; i++)
{
// values in the vector
temp[i].first = arr[i];
// index of the particular value.
temp[i].second = i;
}

``````//sort the temp vector according to the values
sort(temp.begin(), temp.end());
// variable to store the answer
int minimum_swaps = 0;
int i = 0;
while (i < n)
{
// If there is no need to swap then continue
if (temp[i].second == i or temp[i].first == arr[i])
{
++i;
continue;
}
else
{
swap(temp[i].first, temp[temp[i].second].first);
swap(temp[i].second, temp[temp[i].second].second);
if (temp[i].second != i)
i--;
}
minimum_swaps++;
// move to the next index
++i;
}
return minimum_swaps;
``````

}

int main()
{
vector arr = {1, 4, 3, 2};
int n = arr.size();
cout << "Minimum number of swaps required: " << findMinSwap(arr, n) << â€˜\nâ€™;
}

Hope this helps

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